Question

Show that the given nonlinear differential equation is exact. (Some algebraic manipulation may be required. Also, recall the remark that follows Example 1.) Find an implicit solution of the initial value problem and (where possible) an explicit solution. \(\left(e^{t+y}+2 y\right) y^{\prime}+\left(e^{t+y}+3 t^{2}\right)=0, \quad y(0)=0\)

Short answer

The given differential equation is \((e^{t+y}+2 y)y'+(e^{t+y}+3 t^2)=0\). We verified that this equation is exact and found an implicit solution: \(F(t, y) = e^{t+y}+t^{3}+y^{2}-1 = 0\). Unfortunately, it is not possible to find an explicit solution for this equation within elementary functions. Nevertheless, we could apply the given initial condition \(y(0) = 0\), which allowed us to determine that the constant k is -1.

Step-by-step solution

Introducing the given differential equation##

The given differential equation can be written as: \((M(t, y) + N(t, y)y^{\prime}) = 0\) Here, \(M(t, y) = e^{t+y}+3 t^{2}\) and \(N(t, y) = e^{t+y}+2 y\).

Verifying if the differential equation is exact##

To verify that the given differential equation is exact, we need to check if: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\) Calculating \(\frac{\partial M}{\partial y}\): \(\frac{\partial M}{\partial y} = \frac{\partial}{\partial y}(e^{t+y}+3 t^{2}) = e^{t+y}\) Calculating \(\frac{\partial N}{\partial t}\): \(\frac{\partial N}{\partial t} = \frac{\partial}{\partial t}(e^{t+y}+2 y) = e^{t+y}\) Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial t}\), the differential equation is exact.

Finding the implicit solution##

As the equation is exact, there exists a function F(t, y) such that: \(\frac{\partial F}{\partial t} = M(t, y) = e^{t+y}+3 t^{2}\) and \(\frac{\partial F}{\partial y} = N(t, y) = e^{t+y}+2 y\) Integrating \(\frac{\partial F}{\partial t}\) with respect to t, we get: \(F(t, y) = \int (e^{t+y}+3 t^{2}) dt = e^{t+y}+t^{3}+C(y)\), where C(y) is the integration constant with respect to y. Now, differentiating F(t, y) with respect to y, we get: \(\frac{\partial F}{\partial y} = \frac{\partial}{\partial y}(e^{t+y}+t^{3}+C(y)) = e^{t+y}+C^{\prime}(y)\) Comparing this with \(N(t, y)\), we get: \(C^{\prime}(y) = 2y\) Integrating C'(y) with respect to y, we get: \(C(y) = y^{2}+k\), where k is a constant. So, the implicit solution F(t, y) is given by: \(F(t, y) = e^{t+y}+t^{3}+y^{2}+k\)

Applying the initial condition and finding the explicit solution##

Given the initial condition \(y(0) = 0\). We can plug in t = 0 and y = 0 to find the constant k: \(F(0, 0) = e^{0+0}+0^{3}+0^{2}+k = 1+k\) Since y(0) = 0, F(t, y) = 0, and we get: \(0 = 1+k \Rightarrow k = -1\) Now, the implicit solution can be written as: \(F(t, y) = e^{t+y}+t^{3}+y^{2}-1 = 0\) To find the explicit solution, we need to solve this equation for y. However, this equation is complex and cannot be explicitly solved for y in terms of t. Thus, we have found the implicit solution, but the explicit solution is not possible within elementary functions.

Key concepts

Exact Differential Equations

When we talk about exact differential equations, we're discussing a special class of differential equations that can be written in the form (M(t, y) + N(t, y)y'), where M and N are functions of t (time) and y (the dependent variable). An equation is considered exact if the partial derivative of M with respect to y is the same as the partial derivative of N with respect to t, which is mathematically expressed as (∂M/∂y = ∂N/∂t).

Exact differential equations are significant because they can be solved by finding a potential function F(t, y), which satisfies (∂F/∂t = M(t, y)) and (∂F/∂y = N(t, y)). This method simplifies many complex differential equations into a more manageable form.

Initial Value Problem

An initial value problem is a differential equation accompanied by a specific condition, which typically specifies the value of the unknown function at a given point, known as the 'initial value'. For example, if our differential equation is y'=f(t, y), an initial value could be y(t_0) = y_0, where t_0 is the initial time and y_0 is the value of the function y at that time. Solving an initial value problem involves finding a function y(t) that not only satisfies the differential equation but also passes through the point (t_0, y_0) on the graph of the function.

Implicit Solution

An implicit solution to a differential equation involves a relationship between the independent variable t and the dependent variable y that is not explicitly solved for y. In simpler terms, it's a solution where the function y is still intertwined with t in an equation, and we may not be able to isolate y to express it solely as a function of t. The advantage of implicit solutions is that they can often represent a wider range of possible solutions, including those that might be difficult or impossible to express explicitly.

Partial Derivatives

Partial derivatives are a fundamental tool in dealing with functions of several variables like in our case with M(t, y) and N(t, y). They measure how a function changes as only one variable changes while keeping others constant. Mathematically, the partial derivative of a function with respect to a variable is denoted with the script d or the Greek letter δ, such as (∂M/∂y) indicating the partial derivative of M with respect to y while holding t constant. These derivatives are crucial in the verification of exactness and in finding an implicit solution of a differential equation.